poj1017(Packets)

Description

A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square parcels of the same height h as the products have and of the size 6*6. Because of the expenses it is the interest of the factory as well as of the customer to minimize the number of parcels necessary to deliver the ordered products from the factory to the customer. A good program solving the problem of finding the minimal number of parcels necessary to deliver the given products according to an order would save a lot of money. You are asked to make such a program.

Input

The input file consists of several lines specifying orders. Each line specifies one order. Orders are described by six integers separated by one space representing successively the number of packets of individual size from the smallest size 1*1 to the biggest size 6*6. The end of the input file is indicated by the line containing six zeros.

Output

The output file contains one line for each line in the input file. This line contains the minimal number of parcels into which the order from the corresponding line of the input file can be packed. There is no line in the output file corresponding to the last ``null'' line of the input file.

Sample Input

0 0 4 0 0 1 
7 5 1 0 0 0 
0 0 0 0 0 0

Sample Output

2 
1

装箱问题，不难写，就是写起来很麻烦，看了大牛的博客后才把代码减短到这么少的。。。详细题解见注释

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double EPS = 1e-8;
const int MAXN = 50010;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};

int main()
{
    int a, b, c, d, e, f, u[] = {0, 5, 3, 1};//u表示3*3盒子分别为4倍数+0、+1、+2、+3时多出的2*2个数
    while (1)
    {
        scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f);
        if (!a && !b && !c && !d && !e && !f)
            break;
        int ans = d + e + f + (c + 3) / 4;//装3*3、4*4、5*5、6*6需要箱子数
        int y = 5 * d + u[c%4];//装完3*3、4*4多出的2*2个数
        ans += (max(b-y, 0) + 8) / 9;//多出的2*2盒子需要的箱子数
        int x = 36 * ans - 36 * f - 25 * e - 16 * d - 9 * c - 4 * b;//多出的1*1个数
        ans += (max(a-x, 0) + 35) / 36;//多出的1*1盒子需要的箱子数
        cout << ans << endl;
    }
    return 0;
}

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